Muons created 12.0 km above sea level travel downward at 0.998c. Their mean proper lifetime is 2.20 μs. Calculate the Lorentz factor, mean lifetime and mean travel distance in Earth's frame, and the atmosphere thickness in the muon frame. Estimate the survival fraction to sea level using exponential decay in either frame. Explain how the two descriptions agree without assigning an absolute time dilation.

Question

Accepted Answer

γ = 15.82; Earth-frame lifetime = 34.8 μs and mean distance = 10.4 km; muon-frame atmosphere thickness = 0.759 km. Survival fraction ≈ exp(−12.0/10.4) = 0.316. Lorentz factor is γ = 1/√(1−0.998²) = 15.82. The muon lifetime is proper in its own rest frame. Earth observers measure dilated mean lifetime γτ₀ = (15.82)(2.20 μs) = 34.8 μs and mean travel distance vγτ₀ = 10.4 km. The survival probability over 12.0 km is exp[−t/(γτ₀)] with t = 12.0 km/(0.998c), giving exp(−12.0/10.4) = 0.316. In the muon frame, atmospheric thickness is length-contracted: L' = 12.0 km/15.82 = 0.759 km. Travel time L'/(0.998c) compared with τ₀ gives the same exponential factor. Earth describes a longer muon lifetime; the muon describes a shorter atmosphere. Lorentz transformations connect these frame-dependent intervals, so neither frame is an absolute privileged description and both predict the same arrival events.

Muon survival and reference frames

What a complete response should do

How did this one feel?